Let $ heta \in R$ such that $3 \sinh (2 heta)=13-3 e^{2 heta}$, then $ heta=$

Question

Let $	heta \in R$ such that $3 \sinh (2 	heta)=13-3 e^{2 	heta}$, then $	heta=$

Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { } 3 \sinh (2 	heta)=13-3 e^{2 	heta} \ & 3\left(\frac{e^{2 	heta}-e^{-2 	heta}}{2}ight)=13-3 e^{2 	heta} \end{aligned} $ Let $x=e^{2 	heta}$ $\begin{aligned} & \frac{3}{2}\left(x-\frac{1}{x}ight)=13-3 x \ \Rightarrow & 3 x^2-3=2 x(13-3 x) \Rightarrow 3 x^2-3=26 x-6 x^2 \ \Rightarrow & 9 x^2-26 x-3=0 \Rightarrow 9 x^2-27 x+x-3=0 \ \Rightarrow & 9 x(x-3)+1(x-3)=0 \Rightarrow(9 x+1)(x-3)=0 \ \Rightarrow & x=3 \left[\because x 
eq-\frac{1}{9}ight] \ 	herefore & e^{2 	heta}=3 \ \Rightarrow & 2 	heta=\log 3 \Rightarrow 	heta=\frac{1}{2} \log 3\end{aligned}$

Let $\theta \in R$ such that $3 \sinh (2 \theta)=13-3 e^{2 \theta}$, then $\theta=$

Solution